i hate masteringphysics, i perfer doing hw and turning it in so teachers can tell you what your doing wrong.
"Mastering Physics here too, though I'm using Physics with Modern Physics for Scientists and Engineers, Third Edition by Wolfson and Pasachoff."
same here, what hw# are you on...
i had a typo while changing latex into normal text. it's 1500m from C to B. sorry about that.
how would i find the overall velocity of the swimmer? do i need to use the final velocity formula(v(t) = V(0) + at)?
if the horizontal velocity is 5km, wouldnt it help the swimmer to get to points...
A swimmer wants to cross a river, from point A to point B. The distance d_1 (from A to C) is 200 m, the distance d_2 (from C to B) is 1500 m, and the speed V_r of the current in the river is 5 km/h. Suppose that the swimmer makes an angle of 45 degrees(0.785 radians) with respect to the line...
lol, i just figured out what you tried to tell me, im a bit slow. wow, that helped alot. my teacher goes really fast, and doesnt explain the little details and the book is just weird. i already solved 6 problems on my own(14 more to go) and got them all right. thank you agian, i love this forum
sorry, you guys are right....
well anyway, with the correct results...
\frac{tan\theta}{16} + \frac{tan^3\theta}{48} + C
time to sub in theta...
\frac{\frac{\sqrt{(4x^2-1)}}{x^3}}{16} + (\frac{\frac{\sqrt{(4x^2-1)}}{x^3})^3}{48} + C
is that right? i drew a triangle and...
\frac{1}{16} [\int sec^2\theta d\theta + \frac{1}{16}\int sec^2\theta tan^2\theta d\theta]
my work:
found the anti-derv.
\frac{1}{16}(tan \theta) + \frac{1}{16}(sec \theta)^2
ok now this is the part im confused on... i know i need to draw a triangle and do something with it...
dont take my word for it(im not good at physics, lol), but this is how i think your suppose to do it:
x=590 (east)
y=-635 (south, cause it's diving and negative because it's south)
mag is just a^2+b^2=c^2
so... 590^2 + (-635)^2 = c^2
and direction is... arctan(y/x) = arctan...
ok so the problem should look like \int \frac{x^3}{tan \theta} right now.
i dont know what to do next. i cant use u-du. can i use parts on this?
if you set 2x= sec \theta
then x= \frac{sec \theta}{2}
then would i have to take the derv. of that?
dx= 1/2*ln(sec(\theta)+tan(\theta))...